4. If d divides i×n! + 1 and j×n! + 1, then d also divides their difference, which is (j–i)×n!. But all the prime factors of the product (j–i)×n! must lie between 1 and n. It follows that any prime factor of d must lie between 1 and n. So d cannot divide i×n! + 1. Therefore d has no prime factors, so d must be 1.